In past weeks, the average number of life insurance policies sold per week was $θ = 3$. However, this week he has sold 6 life insurance policies. Based on this single observation of $x = 6$, what is the value of the likelihood-ratio test statistic for testing the hypothesis that the salesman has suddenly seen a boost in his job, i.e. testing $H_0 = \theta = 3$ , $H_1 = \theta > 3$.

For this, I used the likelihood ratio for poisson distribution, giving me $$\frac{\frac{3^6e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}}$$ This gives 0.3138365. Is this the right answer?

  • $\begingroup$ Your fraction reduces to $3^3/6^6 = .5^6= 0.015625.$ $\endgroup$ – BruceET Apr 29 '19 at 19:52
  • 2
    $\begingroup$ @BruceET - thought the fraction was probably intended to be $\frac{\frac{3^6 e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}} =\dfrac{\exp(3)}{2^6}$ $\endgroup$ – Henry Apr 29 '19 at 20:01
  • 1
    $\begingroup$ Yes that was the intended fraction, just typed it wrong. Thank you for clarifying! $\endgroup$ – martinhynesone Apr 29 '19 at 20:15

You will want to reject $H_0: \theta = 3$ against $H_a: \theta > 3)$ for large values of your single observation $X.$ You can't test at exactly the 5% level because of the discreteness of the Poisson distribution.

Under $H_0,$ you have $P(X \ge 7) = .0335,$ but $P(X \ge 6) = .0834.$ Exact Poisson computations from R (where ppois is a Poisson CDF and qpois is the quantile function):

qpois(.95, 3)
[1] 6
1 - ppois(6,3)
[1] 0.03350854  # 1 - P(X <= 6) = P(X >= 7)
1 - ppois(5,3)
[1] 0.08391794

So for test at level $\alpha =3.335,$ you should reject if $X \ge 7.$ Because you observe $X = 6,$ the P-value is $P(X \ge 6\,|\,\theta = 3) = 0.834.$

In the plot below, $\alpha$ is the sum of the heights of the bars to the right of the vertical dotted line.

enter image description here

For such a small values of $\theta,$ depending on the desired accuracy, it may no bet advisable to use a normal approximation. The best-fitting normal curve is shown in blue.

However, using the normal approximation (with continuity correction), the P-value is 7.4%, so you would not reject at the 5% level.

1 - pnorm(5.5, 3, sqrt(3))
[1] 0.07445734

Addendum: Now that you have the correct likelihood ratio function $\lambda(x) = (3/x)^xe^{x-3}$ for $x \ge 3 \; (1$ for $x < 3),$ let's use R to plot it. Then we can see that $\lambda(x)$ is small (leading to rejection) for large values of $x$ as mentioned at the start.

x = 3:10;  like = (3/x)^x*exp(x - 3)
plot(x, like, pch=20, main="Likelihood Ratio")
abline(h = 0, col="green2")

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.